# extrema on a closed curve in $\mathbb R^3$

Let $S$ be the surface in $\mathbb R^3$ given by the equation $z=\frac12 x^2+\frac12 y^2$ and let $R\subset R^3$ be the surface given by $y^2+z^2=3$ ($R$ is the boundary of the infinite cylinder around the $x$-axis). Let $C$ be the closed curve defined as the intersection between $S$ and $R$. I want to find the maximum and minimum of $f(x,y,z)=x^2+y^2+(z-1)^2$ on $C$.

How can I do this? I've tried using the Lagrange multipliers, but it is quite a lot of work.

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Can you use the fact that, on the surface $S$, $2z = x^2+y^2$? Then, on the surface, $f(x,y,z) = z+(z-1)^2$, which you can easily maximize wrt $z$. You still need a way to use the second constraint, but I'm not quite sure what the second constraint is. What do you mean by "the surface given by $y^2+z^2$"? – James Fennell Jul 28 '12 at 10:25
I made a typo, thank you for pointing it out – kevin Jul 28 '12 at 10:27
If you succeed in finding a parametrization $c(t)$ of the intersection curve, then $f\circ c$ is a function $\mathbb{R}\rightarrow \mathbb{R}$. – user20266 Jul 28 '12 at 11:11

James' observation simplifies the problem considerably: In all points of $S$, whence of $C$, we have $$f(x,y,z)=z^2-z+1=\Bigl(z-{1\over2}\Bigr)^2+{3\over4}\ .$$ Therefore we only have to control the value of $z$ along $C$.

The curve $C$ is a sling which encircles the $z$-axis in the $z>0$ part of ${\mathbb R}^3$. As $z=\sqrt{3-y^2}$ the value $z$ is maximal on $C$ when $y=0$, and we have $z_{\rm max}=\sqrt{3}$. Similarly, the minimal value of $z$ will be assumed in the plane $x=0$. Here we have $2z=y^2=3-z^2$; so $z_{\min}=1>{1\over2}$.

It follows that $f\restriction C$ is maximal in the two points $(\pm x_0 ,0,\sqrt{3})$, where $x_0$ is determined by the equation $2z=x^2$, and $f\restriction C$ is minimal in the two points $(0,\pm\sqrt{2},1)$.

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If this is what it looks like (namely, homework), then I think you are expected to use Lagrange with two constraints. Yes, that is five equations, but three of them are linear, and if you solve them first you may be able to do the quadratics later.

A smarter approach is to use the constraints to simplify the target function. From the second you'll get $f=x^2-2z+\mathrm{const}$, and then from the first $f=-y^2+\mathrm{const2}$. This clears up things a bit, especially with the form of the second constraint.

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